Let F_1 and F_2 be the foci of an ellipse fra | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For an ellipse,the reflected ray from one focus passes through the other focus.Therefore,the incident ray is $PF_1$ and the reflected ray is $PF_2

Vedclass · Accepted Answer

$y = x + \frac{5}{\sqrt{13}}$

Vedclass · Answer

(A) For an ellipse,the reflected ray from one focus passes through the other focus.Therefore,the incident ray is $PF_1$ and the reflected ray is $PF_2$.The angle bisector of the angle between the incident ray $PF_1$ and the reflected ray $PF_2$ is the normal to the ellipse at point $P$.The equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{9} = 1$,where $a^2 = 4$ and $b^2 = 9$.Since $b^2 > a^2$,the major axis is along the $y$-axis.The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \frac{(a^2 - b^2)m}{\sqrt{a^2 + b^2m^2}}$.Substituting $a^2 = 4$,$b^2 = 9$,and $m = 1$ (assuming the bisector has slope $1$ as per options):$y = x \pm \frac{(4 - 9)(1)}{\sqrt{4 + 9(1)^2}} = x \pm \frac{-5}{\sqrt{13}}$.Thus,the equation is $y = x \pm \frac{5}{\sqrt{13}}$.

Let $F_1$ and $F_2$ be the foci of an ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$. $A$ ray from $F_1$ strikes the elliptical mirror at point $P$ and is reflected. What is the equation of the angle bisector of the angle between the incident ray and the reflected ray?

Similar Questions

The coordinate of a point on the auxiliary circle of the ellipse $x^{2}+2y^{2}=4$ corresponding to the point on the ellipse whose eccentric angle is $60^{\circ}$ will be

The least intercept made by a tangent to the ellipse $\frac{x^2}{64}+\frac{y^2}{49}=1$ with the coordinate axes is

For the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$,a chord $PQ$ subtends a right angle at the center. What is the locus of the point of intersection of the tangents at $P$ and $Q$?

Find the coordinates of the foci,the vertices,the length of the major axis,the minor axis,the eccentricity,and the length of the latus rectum of the ellipse $\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$.

The lengths of the sides of the rectangle of greatest area that can be inscribed in the ellipse $x^2+4y^2=64$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module